A069211 Let u(n,k) be the recursion: u(n,1)=1, u(n,2)=n, u(n,k+2) = (1/2) * (u(n,k+1)+u(n,k)) if u(n,k+1)+u(n,k) is even, and u(n,k+2) = abs(u(n,k+1)-u(n,k)) otherwise. Sequence gives integer values a(n) such that u(n,k)=1 for any k>=a(n).

1, 3, 4, 6, 8, 7, 7, 9, 13, 11, 8, 10, 16, 10, 10, 12, 14, 16, 12, 14, 14, 11, 11, 13, 15, 19, 11, 13, 17, 13, 13, 15, 21, 17, 17, 19, 18, 15, 15, 17, 21, 17, 12, 14, 19, 14, 14, 16, 24, 18, 20, 22, 22, 14, 14, 16, 19, 20, 14, 16, 25, 16, 16, 18, 20, 24, 18, 20, 28, 20, 20, 22

Offset

1,2

Comments

It seems that Sum_{i=1..n} a(i) ~ C*n*log(n) asymptotically with C=0.2...

Links

Table of n, a(n) for n=1..72.

Sean A. Irvine, Java program (github)

Formula

a(2n) = a(n) + 2 + [n>1]. - Ralf Stephan, Oct 08 2003

Example

Let n=7, for k=1,2,3,4,5,6,7,8 u(7,k)=1,7,4,3,1,2,1,1 hence a(7)=7 since for all k>=7 we have u(7,k)=1.

Crossrefs

Sequence in context: A225794 A206769 A141219 * A368152 A242822 A295996

Adjacent sequences: A069208 A069209 A069210 * A069212 A069213 A069214

Keyword

nonn

Author

Benoit Cloitre, Apr 11 2002

Extensions

Name corrected by Sean A. Irvine, Apr 12 2024

Status

approved